In-span loads on beam elements So far we have only been able to apply loads at nodes. example How do...

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Transcript of In-span loads on beam elements So far we have only been able to apply loads at nodes. example How do...

In-span loads on beam elements

So far we have only been able to apply loads at nodes.

example

How do we then tackle loads away from nodes, or continuous loads?---------------------------------------------------------------Option 1: add a node.

example

Option 2: fool the structure into thinking it has in-span loads when it doesn’t. This is the (more powerful) technique we will study in detail.

We have to apply loads/moments at nodes which have the same effect on the structure as the in-span loads.

We conduct two analyses

(A)Clamp all the nodes in the structure, apply the in-span loads and work out the reactions at the clamps.

(B)Release the clamps, remove the in-span loads and reverse the reactions at nodes and conduct a standard matrix stiffness method analysis of the structure.

Why does this work?

It is the principle of superposition which relates to linear elastic structures only.

It does not matter in which order you apply loads to a structure, the deflections/rotations will be the same

An example – a propped cantilever

P Deflected shape under load ( this is what we want to find out)

P

Analysis A

Analysis B

Fixed end moments and

forces

So that we do not ever have to carry out analysis A we use tables of “answers” from A type analyses.

Example – why can we get away with this?

Fixed end moments/reactions are available for a number of load cases (tables in many textbooks) one is on DUO.

Can you derive them? (Yes).

udl of w/m

L

Properties: E, I, A

An example – cantilever with a UDL

We know that:End deflection =

End slope =

Vertical support reaction =

Moment support reaction =

EI

wL

8

4

EI

wL

6

3

wL

2

2wL

1 21

1

1

V

U

2

2

2

V

U1

1

1

M

F

F

Y

X

2

2

2

M

F

F

Y

X

2

2

2

23

2 46

612

12

2V

L

EI

L

EIL

EI

L

EI

wL

wL

We can ignore axial effects as well as those d.o.f.s which are fixed

P

udl of w/m

2L

L2L/3

A

B C

D

Node nos.

PFE effects example 2: Portal frame

Properties

E,I, A

45o

2

1

3

4

P

udl o f w /m

2 L

L2 L /3

A

B C

D

N ode nos.

PPortal frame example

“first load” vector, i.e. those loads already at nodes

2

1

3

4

00

00 0 01XF 1YF 1M 4XF 4YF 4M2P

2P

Nodal load vector (transposed)

Analysis A

27

4PL

27

2PL

27

20P

27

7P

wL wL

3

2wL

3

2wL

wL wL

3

2wL

3

2wL

27

4PL

27

2PL

27

20P

27

7P

Node 3

Node 4Node 1

Node 2

FE effects force vector

27

2PL 27

7PNode 1

00

FE effects force vector

27

7P0

27

2PL

wL

3

2wL

27

4PL

27

20P

Node 2

00

27

7P0

27

2PL27

20PwL

327

4 2wLPL

wL

3

2wL

Node 3

Node 4

00

27

7P0

27

2PL27

20PwL

327

4 2wLPL

wL3

2wL0

Node 4

00

27

7P0

27

2PL27

20PwL

327

4 2wLPL

wL3

2wL0 0 0 0

00

27

7P0

27

2PL27

20PwL

327

4 2wLPL

wL3

2wL0 0 0 0

Bringing the two load vectors together

“first load” vector, i.e. those loads already at nodes

00

00 0 01XF 1YF 1M 4XF 4YF 4M2P

2P

So what we now have to solve is this

4

4

4

3

3

3

2

2

2

1

1

1

4

4

4

2

2

1

1

1

matrix Stiffness Global1212

3

2

2

327

4

2720

272

277

V

U

V

U

V

U

V

U

M

F

F

wLwLP

P

wLPLwL

P

PLM

F

PF

Y

X

Y

X

FE effects example 3: 2 span beam

Error lurking

udl of w/m

L

Results of stiffness matrix

analysis

"Free" structure

"Free" BMD

Final BMD

This is a propped cantilever

In-span bending moments